# Abscissa of convergence for a very specific Dirichlet series / Euler product

I am interested in the convergence of the following Euler product: $$\prod_p \frac{1}{1-\chi(p)\cdot p^{-s}}.$$ The product is over all primes (in increasing order), with $$\chi(p)=+1$$ if $$p \bmod 4 =3$$ and $$\chi(p)=-1$$ if $$p \bmod 4 =1$$. Here $$\chi(2)=\pm 1$$, the sign does not matter. Also $$s=\sigma+ it$$ to use the standard notation.

Since the densities, for both set of primes, are identical, the signs will alternate nicely on average, and one would guess that the product converges not only for $$\sigma>1$$, but actually for $$\sigma>0$$. Is this the case, or is this still a conjecture?

Also, I'd love to see a plot of the orbit for a fixed value of $$\sigma$$, say $$\sigma=0.75$$. By orbit, I mean a scatterplot of the product in the complex plane (real vs. imaginary part), for a fixed value of $$\sigma$$, and $$t$$ varying between $$0$$ and (say) $$1000$$. See example here and here for a truncated (finite) product when $$\chi$$ is constant equal to $$1$$.The hole in the orbit shrinks to a single point (the origin) if you add all the primes [if RH if true; if not it shrinks to an empty set].

I'm working on a number theory tutorial, and I'd like to introduce the students to a situation where the orbit not only never crosses the origin (as for the standard Riemann Hypothesis if $$0.5<\sigma<1$$), but in addition, stays away and never gets too close to the origin. If possible, not a conjectured result, but an established one.

On a side not, I am interested to know what the abscissa of convergence is, for the associated Dirichlet series.

Update - Attempt with a different $$\chi$$

Let $$p_k$$ be the $$k$$-th prime ($$p_1=2$$ and so on) and $$\chi(p_{2k})=-1$$, $$\chi(p_{2k+1})=+1$$. Let $$\chi$$ be completely multiplicative. Could the abscissa of convergence be strictly less than $$1$$? If you look at the Euler product, say for $$\sigma=0.75$$, its convergence at $$t=0$$ is equivalent to the convergence (after taking the logarithm and standard asymptotic) of $$\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{p_k^\sigma}.$$ I believe that the Dirichlet test could positively answer the question. Actually, there was a question asked on MO eight years ago [see here], about a similar convergence problem (proved to converge) so similar that this one must also converge if $$\sigma=0.75$$, and indeed, as long as $$\sigma>0$$.

Assuming this is correct, and since convergence for a specific $$\sigma_0=0.75$$ at $$t=0$$ implies convergence for all $$t$$ for that particular $$\sigma_0$$, it implies that the abscissa of convergence is well below $$1$$. Does this also hold for the Euler product? If yes (and I assume that the answer is yes), then can the product still be equal to zero? The norm of the product can be very close to zero. My guess is that for the product to be zero if $$\sigma<1$$ (that is, to diverge) you would need $$t=\infty$$. I could be wrong.

Now defining the concept of hole. In dynamical systems, it's called a repulsion basin, usually not circular. Let $$L(z,\chi)$$ be the Dirichlet function represented by the product or its associated series. Here the hole is centered at the origin $$z_0=(0,0)$$ in the complex plane. It is the largest circle of radius $$\rho$$ such that $$|L(z,\chi)-z_0|>\rho$$ for all $$z=\sigma+it$$ in the complex plane, for a fixed value of $$\sigma$$, say $$0.75$$.

There is one thing where I am most certainly wrong: the hole is reduced to to a single point (the origin), contrarily to what I thought initially (a hole of radius $$\rho>0$$). But if you make a video of the orbit starting at $$t=0$$, no matter how fast your computer is, for a very, very long time the hole will be visible to the naked eye. It will shrink as $$t$$ increases, in the end to a single point (or worse, an empty set if there are some unexpected roots) but incredibly slowly. That's what it does for the standard RH case.

Final update: non-trivial examples with a hole

This does not contradict the universality property: this property is true if $$\sigma<1$$ and applies if $$P$$ is the set of all primes. Here $$\sigma<1$$ (typically but not necessarily) and $$P$$ can be a subset of primes, finite or infinite. A modified (generalized) universality property, in this context, would be: the orbit is dense between its internal boundary (the boundary of the hole), and its external boundary if bounded (e.g. if $$\sigma>1$$). Still have to prove it. Also, I assume that if there is a "thick" hole, there is only one.

I could not find "standard" examples (characters modulo $$m$$) with a hole if $$P$$ contains sufficiently many primes. I am pretty sure there are no such examples. So you need to look at cases where $$P$$ is infinite but sparse enough. It will work (still have to go through a round of double-checking to confirm this) if

$$\rho=\prod_{p\in P} \frac{1}{1+p^{-s}}>0$$

regardless of $$\chi$$. In other words, the above product needs to converge. Then the radius of the hole is $$\geq \rho$$. Using my generalized version of the universality property, I guess the radius is $$\rho$$ (exactly). This is compatible with RH ($$\rho=0$$) and GRH (in that case $$P$$ is too big anyway, and $$\rho=0$$).

Examples with a hole with $$\rho>0$$:

• If $$P$$ is the set of all primes that are the sum of two cubes.
• If $$P$$ is the set of twin primes ?? (this case is borderline)
• You write "since convergence for a specific $\sigma_0 = 0.75$ at $t=0$ implies convergence for all $t$ for that particular $\sigma_0$, it implies that the abscissa of convergence is well below 1". Be careful: convergence of a Dirichlet series at one complex number $s_0$ implies its convergence at all $s$ with ${\rm Re}(s) > {\rm Re}(s_0)$, but not convergence everywhere on the line ${\rm Re}(s) = {\rm Re}(s_0)$. For example, the series over primes $\sum_p 1/p^s$ converges for all $s$ with ${\rm Re}(s) = 1$ except at $s = 1$, which it is infinite. Jul 3 at 2:20
• The totally multiplicative function $\chi$ in your update seems rather artificial since it amounts to assigning to a prime the value $1$ or $-1$ based on it being an even-indexed or odd-indexed prime, but I don't recognize that as being related to anything that's already known (in contrast to looking at primes based on their residue mod $m$ for some $m$). That series $\sum \chi(p)/p^s$ converges for ${\rm Re}(s) > 0$ since the partial sums of its coefficients in their natural order are bounded (as they only take the values $\pm 1$ and $0$), but I don't see how it tells us anything. Jul 3 at 2:27
• @KConrad: I still have an issue with my product. If $\chi=1$ is constant (standard RH) the norm of the product is bounded by $\prod (1+p^{-\sigma})^{-1}$. This can be zero: it does not help. Of course product makes no sense if $\sigma\leq 1$. But in my question here, the product converges (at least in my update) for $\sigma=0.75$ (say). At least it convergences if $t=0$. But does it converge if $t\neq 0$? All I could get is that the norm of the product is again bounded by $\prod (1+p^{-\sigma})^{-1}$. This does not help. Jul 3 at 2:34
• The Euler product for $L$-functions of nontrivial Dirichlet characters, with terms indexed by primes in the usual order, converges on ${\rm Re}(s) = 1$ to the $L$-function of the Dirichlet character. Your $\chi$, with values depending on a prime being in an even or odd place, is an awkward thing. Your goal is to present something to students, so show them conventional concepts that are widely used instead of peculiar constructions when it is hard to see how they behave. If a widely used concept has a subtlety, at least it is known to be important. But if something strange is subtle, so...? Jul 3 at 3:59
• About the "hole": I hope you are aware of the universality theorems for Dirichlet L-functions. See in particular "3.3. Corollary" in eudml.org/doc/173236 and apply it to a singleton set D={z}. Jul 3 at 10:54

What do you mean by the word "orbit"? Please define that term in the body of your question.

You ask where the product over $$p$$ converges. Although you did not specify it, I assume you mean product with terms in the order of increasing $$p$$. You need to specify the order of the terms if you're dealing with something not necessarily absolutely convergent.

Your $$\chi$$ is meant to be a totally multplicative function. For odd primes $$p$$, your $$\chi(p)$$ is $$\lambda(p)\chi_4(p)$$ where $$\lambda$$ is the Liouville function (the nonvanishing totally multiplicative cousin of the Moebius function that is $$-1$$ at all primes) and $$\chi_4$$ is the nontrivial Dirichlet character mod $$4$$, so a calculation with Euler products shows for $${\rm Re}(s) > 1$$ that (as GH from MO points out) $$\sum_{n \geq 1} \frac{\chi(n)}{n^s} = \frac{1}{1-\chi(2)/2^s}\frac{L(2s,\chi_4^2)}{L(s,\chi_4)} = \left(1 + \frac{\chi(2)}{2^s}\right)\frac{\zeta(2s)}{L(s,\chi_4)}.$$ The Euler product for $$\zeta(2s)$$ converges absolutely for $${\rm Re}(s) > 1/2$$ and $$1 + \chi(2)/2^s$$ is nonvanishing for $${\rm Re}(s) > 0$$, so for $${\rm Re}(s) > 0$$ your Euler product will converge exactly where $$\prod_p (1 - \chi_4(p)/p^s)$$ converges.

The Generalized Riemann Hypothesis for $$L(s,\chi_4)$$ implies for $${\rm Re}(s) > 1/2$$ that $$L(s,\chi_4) \not= 0$$ and that $$L(s,\chi_4) = \prod_p 1/(1-\chi_4(p)/p^s)$$. Conversely, the convergence and nonvanishing of $$\prod_p 1/(1-\chi_4(p)/p^s)$$ at a number $$s_0$$ with $${\rm Re}(s_0) > 1/2$$ implies it equals $$L(s_0,\chi_4)$$ and that $$L(s,\chi_4) \not= 0$$ for $${\rm Re}(s) > {\rm Re}(s_0)$$. So it is not reasonable to expect the product to converge for $${\rm Re}(s) < 1/2$$ and it is reasonable to expect the product to converge for $${\rm Re}(s) \geq 1/2$$ excluding at zeros of $$L(s,\chi_4)$$ on the critical line, but the only way you're going to have access to such results is through GRH.

It is utterly hopeless to expect a proof of convergence of either $$\prod_p 1/(1-\chi_4(p)/p^{s_0})$$ or $$\sum \chi_4(p)/p^{s_0}$$ at a single $$s_0$$ where $${\rm Re}(s_0) < 1$$ unless you expect to be making dramatic progress towards GRH for $$L(s,\chi_4)$$, and the same is true with $$\chi_4$$ replaced by $$\chi$$. Concerning the Dirichlet series $$\sum \chi(n)/n^s$$, since $$\chi(n) = \lambda(n)\chi_4(n)$$ for odd $$n$$ it's reasonable to expect $$\sum_{n \leq x} \chi(n) = O_\varepsilon(x^{1/2+ \varepsilon})$$ if you want to use GRH for $$L(s,\chi_4)$$, so it's reasonable to expect that $$\sum \chi(n)/n^s$$ converges for $${\rm Re}(s) > 1/2$$, but it's not reasonable to expect a proof of that without a major advance on GRH. For an analogous situation where a Dirichlet series was proved to converge on a half-plane bigger than its half-plane of absolute convergence by relying on a major result (modularity of elliptic curves over $$\mathbf Q$$), see my answer to an earlier MO question here.

Watch out: there is probably weird behavior for the Euler product on the critical line. We have $$L(1/2,\chi_4) \not= 0$$ (not weird) and if $$\prod_p 1/(1-\chi_4(p)/\sqrt{p})$$ has a nonzero value, then that value is $$\sqrt{2}L(1/2,\chi_4)$$ (weird!). See my paper "Partial Euler products on the critical line" or Kuo and Murty's paper "On a conjecture of Birch and Swinnerton-Dyer". Both appeared in Canadian J. Math. in 2005.

Since your purpose in working with the Euler product $$\prod_p 1/(1-\chi(p)/p^s)$$ is purely to have a concrete example for some students, I suggest you work with $$\chi_4$$, not $$\chi$$: the function $$\chi_4$$ is a more commonly encountered object (a Dirichlet character), $$\chi_4$$ doesn't have an artificial definition at $$p = 2$$, and the half-plane of convergence of the Dirichlet series $$\sum \chi_4(n)/n^s$$ is already known to be precisely $${\rm Re}(s) > 0$$.

• I will update my question accordingly. The orbit is a plot, in the complex plane, of the product, for a specific $\sigma$ and varying $t$, say $t\in[0, T]$ with $T=200$. Jul 2 at 19:34
• @ KConrad: thank your your answer. I'll read your paper, especially since I worked quite a bit with finite Euler products. Here's a new question: Let $p_k$ be the $k$-th prime. If you choose $\chi(p_{2k})=-1, \chi(p_{2k+1})=+1$ then wouldn't things be easier? Wouldn't you get a value well below $1$, for the abscissa of convergence? This MO answer suggests that it does: mathoverflow.net/questions/159534/… Jul 3 at 0:48

The OP's Euler product can be written as $$\bigl(1+\chi(2)2^{-s}\bigr)\frac{\zeta(2s)}{L(s,\chi_4)},$$ where $$\chi_4$$ is the nontrivial Dirichlet character mod $$4$$. As a result, for any $$\sigma\in(1/2,1)$$, the convergence of the OP's Euler product in the half-plane $$\Re s>\sigma$$ is equivalent to the non-vanishing of $$L(s,\chi_4)$$ in the same half-plane. In particular, the abscissa of convergence is less than $$1$$ if and only if a quasi Riemann Hypothesis holds for $$L(s,\chi_4)$$. The quasi Riemann Hypothesis is a major open problem for any Dirichlet $$L$$-function (or any automorphic $$L$$-function).

• At least, if $s$ is real, do we know if the product converges for some $\sigma_0<1$? Taking the logarithm, this is equivalent to proving that $\sum_p \chi(p)p^{-\sigma_0}$ converges. If the terms were perfectly alternating, the series would converge even for $\sigma_0=0.001$. Jul 2 at 19:30
• @VincentGranville If the product converges for some $s_0=\sigma_0+it_0$ with $\sigma_0>1/2$, then it also converges in the half-plane $\Re s>\sigma_0$. So convergence for any $s_0$ with $1/2<\Re s_0<1$ is equivalent to a quasi Riemann Hypothesis for $L(s,\chi_4)$. Jul 2 at 20:45
• In the end, what I would like to visually display is stronger than "$\sigma_0>1/2$". It's the fact that there is an actual hole, visible to the naked eye, around the origin, for some $\sigma<1$ (be it $\sigma=0.99$). GRH alone states that yes, there is an hole, but it could be as small as a single point. My belief is that this hole has Lebesgue measure $>0$. I'll tell my students that despite my assumption being so intuitive (at least to me), it hasn't been proved yet. I have no intent to prove or disprove it. Jul 2 at 21:03
• What do you mean by “a hole”: a positive real number at which some series or product does not converge? Jul 2 at 22:13
• Oops, I thought the OP sees all comments anywhere on the page. Jul 2 at 23:54